Question

find K if the sum of roots of polynomial x2-x+k(2x-1)is zero

Answer 1

EXPLANATION.

sum of the polynomial = x² - x + k(2x - 1 ) = 0.

⇒ x² - x + k(2x - 1 ) = 0.

⇒ x² - x + 2kx - k = 0.

⇒ x² - (1 - 2k )x  - k = 0.

To find the value of k.

as we know that,

D = 0  Or b² - 4ac = 0.

⇒ -( 1 - 2k )² - 4(1)(-k) = 0.

⇒ 1 + 4k² - 4k + 4k = 0.

⇒ 4k² + 1 = 0.

⇒ k² = -1/4.

⇒ k = √-1/4.

                     

MORE INFORMATION.

Sign of quadratic expression ⇒ ax² + bx + c.

if α,β are the roots roots of the corresponding quadratic equation then for x = α and x = β the value of the expression ax² + bx + c is equal to 0. for other real value f x, the expression ax² + bx + c > 0 or < 0.

The sign of ax² + bx+ c id determine by the followig rule.

(1) = if α,β ( α < β ) are real and unequal ( i.e D > 0 ) roots of the corresponding quadratic equation then the sign of y = ax² + bx + c, x ∈ R is determine as follows,

(a) = clearly y is +ve for x < α or x > β and y is -ve for α < x < β.

a > 0 , D > 0

(b) = clearly y is +ve for α < x < β and y is -ve for x < α and x > β

a < 0 , D > 0.

Answer 2

Given:-

sum of zeroes = 0

hence, -b/a = 0

Solution:-

x²-x+k(2x-1) = 0

x²-x+2kx-k = 0

x²-(1+2k)x-k = 0

now, $\bf\dfrac{-b}{a}$ = $\bf\dfrac{1-2k}{1}$ = 0

1-2k = 0

1 = 2k

k = $\bf\dfrac{1}{2}$